Answer:
The answer to this question can be defined as follows:
Step-by-step explanation:
In the question equation is missing so, the equation and its solution can be defined as follows:
[tex]B={b_1,b_2}\\\\b_1= \left[\begin{array}{c}5&5\end{array}\right] \ \ \ \ \b_2= \left[\begin{array}{c}2&-5\end{array}\right] \ \ \ \ \x= \left[\begin{array}{c}-7&-35\end{array}\right][/tex]
[tex]\left[\begin{array}{c}a&c\end{array}\right] =?[/tex]
[tex]\to \left[\begin{array}{c}-7&-35\end{array}\right]= a\left[\begin{array}{c}5&5\end{array}\right]+c \left[\begin{array}{c}2&-5\end{array}\right] \\[/tex]
[tex]\to \left[\begin{array}{c}-7&-35\end{array}\right]= \left[\begin{array}{c}5a+2c&5a-5c\end{array}\right]\\\\\to 5a+2c=-7....(1)\\\\\to 5a-5c=-35....(2)\\\\[/tex]
subtract equation 1 from equation 2:
[tex]\to 7c=28\\\\\to c=\frac{28}{7}\\\\\to c= 4\\\\[/tex]
put the value of c in equation 1
[tex]\to 5a+2(4)=-7\\\to 5a+8=-7\\\to 5a=-7-8\\\to 5a=-15\\\to a= -3[/tex]
coordinate value is [-3,4].